Theoretical Background

Let f(t) denote a continuous Graph having a continuous Fourier transform. Let f(t) be sampled at uniform discrete intervals of T pixels. Then, according to Shannon's Sampling Theorem f(t) can be exactly reconstructed from its samples f(n.T) if and only if the spectrum F(ω) of f(t) has a limited bandwith, such that F(ω) = 0 for all |ω| ≥ π/T . It is clear from the start, however, that the sampling interval T of a Graph in an image is just equal to T = 1 pixel .
We seek a Continuation of the discrete Graph which is representative. By this we mean that the continuation can be (re-)constructed from the Discrete background. Hence the bandwith of its Fourier Transform should be such that |ω| ≥ π/T = π .
We will define now a quite a specific continuation f of the Graph, as follows (apart from a well-known constant):
   f(t) = Σ_k f_k e^{-[(t - t_k) / σ]2 / 2}     with discrete values   f_k   at positions   t_k
This kind of continuation may be called a Sense. Here the "spread" σ is still to be determined.
Calculate the Fourier Transform F of the Sense (apart from norming constants again):
   F(ω) = Σ_k f_k e^{i ω t_k} . e^{-ω2 σ2/2}
The result is equal to kind of a Fourier Transform of the sampling, multiplied by a factor which is independent of the sampling as well as the samples. It will be assumed that the whole F(ω) becomes effectively zero (that is: small enough) if this sampling-independent factor becomes smaller than our standard accuracy (-: which can be redefined it if such is not the case):
   e^{-ω2 σ2/2} < e^-(2π)2/2    giving:    |ω σ| > 2π    or:    |ω| > 2 π/σ = π / (σ/2)
But we know from Shannon's Sampling Theorem that: |ω| ≥ π/1 .
Giving for the quantity σ an extremely simple rule of thumb:    1 < σ/2    or    σ > 2 .
In words: the spread of the bell-shapes in sensing the discrete Graph must at least be greater that two times the pixel size.
And the smaller σ , the better, of course. In practice, values 2≤ σ ≤ 3 may be suitable.

The continuation of a discrete background has the immense advantage that it can be manipulated with help of the whole machinery of classical calculus. We repeat:
   f(t) = Σ_k f_k e^{-[(t - t_k) / σ]2 / 2}
This sum of Gauss shapes is much less cumbersome than it seems at first sight, because we have seen that only values at intervals smaller than 2.πσ , where σ has a value, say, between 2 and 3 pixels, have to be calculated.
The function can easily be differentiated, for example:
   f '(t) = - Σ_k f_k [(t - t_k) / σ²] e^{-[(t - t_k) / σ]2 / 2}
Also the second derivative can be calculated without much effort:
   f ''(t) = Σ_k f_k [(t - t_k) / σ²]² e^{-[(t - t_k) / σ]2 / 2} - 1 / σ² Σ_k f_k e^{-[(t - t_k) / σ]2 / 2}
In a nutshell, this is the whole secret behind Numerical Differentiation, as it is accomplished according to my insights (HdB).